/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 During an auto accident, the veh... [FREE SOLUTION] | 91Ó°ÊÓ

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During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, air bags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g} ?\)

Short Answer

Expert verified
The passenger travels approximately 0.38 meters.

Step by step solution

01

Identify Given Values and Units

We are given the maximum acceleration and the time duration during which this acceleration occurs.- Maximum acceleration ( a ) is \( 60g \), where \( g = 9.81 \, \text{m/s}^2 \).- Time duration ( t ) is \( 36 \, \text{ms} = 0.036 \, \text{s} \).
02

Calculate the Acceleration in m/s²

Convert the acceleration from \( g \) units to \( \text{m/s}^2 \) by multiplying with \( 9.81 \, \text{m/s}^2 \).\[ a = 60g = 60 \times 9.81 \, \text{m/s}^2 \]\[ a = 588.6 \, \text{m/s}^2 \]
03

Use the Kinematic Equation for Distance

Use the formula for distance with constant acceleration:\[ d = \frac{1}{2} a t^2 \]Substitute the known values for \( a \) (acceleration) and \( t \) (time) into this equation.
04

Calculate Distance

Substitute the calculated values:\[ d = \frac{1}{2} \times 588.6 \, \text{m/s}^2 \times (0.036 \, \text{s})^2 \]Calculate:\[ d = \frac{1}{2} \times 588.6 \times 0.001296 \, \text{m} \]\[ d \approx 0.3812 \, \text{m} \]
05

Round and Interpret the Answer

The distance traveled during the deployment of the airbag is approximately 0.3812 meters. Since we assume the data given is with standard laboratory precision, we round the answer to 0.38 meters for clarity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Acceleration in Crash Scenarios
When you think about cars coming to a stop, acceleration plays a huge role. Acceleration describes how quickly an object's velocity changes. In the context of car crashes and airbag deployment, the term "acceleration" can often be misunderstood because, during a collision, cars decelerate, or reduce speed.\(60g\) indicates that the car and its passengers experience a deceleration rate 60 times greater than the standard gravitational pull on Earth.This is not something people typically experience, so despite the deceleration, we still call it acceleration.
This high acceleration is why airbags are so important. They are designed to slow the person down gradually compared to hitting a hard surface suddenly. Such gentle deceleration is key to reducing injury during a crash.
Using Kinematic Equations for Safety Calculation
In scenarios like vehicle crashes, kinematic equations help calculate important parameters such as distance or time. Here, the kinematic equations relate distance, initial velocity, time, and acceleration. We use the formula:\[ d = \frac{1}{2} a t^2 \] This equation doesn’t require the initial velocity as the object is stopping completely, implying it started from a certain velocity and reached zero.
By using known values for acceleration and time, we can solve for the distance, enabling us to estimate how much space is needed for the airbag to safely stop a person. This highlights the importance of physics in engineering safety features like airbags.
The Vital Role of Airbag Deployment
Airbags are critical safety components in vehicles. During a collision, they inflate rapidly, cushioning the passengers and preventing direct contact with hard parts of the car like the dashboard or windshield. This rapid inflation decreases the impact force experienced by passengers by increasing the time over which deceleration occurs.
Without airbags, a passenger's body could face the full brunt of a crash immediately against the car's inside surfaces. This increased stopping distance reduces the risk of injury, showcasing how understanding physics and engineering principles protects lives.
Calculating Distance During Airbag Deployment
To understand how far a person moves when an airbag deploys, consider the concept of stopping distance. The stopping distance is not just a function of speed but also time and acceleration. Given a high deceleration rate\(60g\) and a brief duration \(36\,\mathrm{ms}\), kinematics enables us to calculate this critical distance.
Using the equation \( d = \frac{1}{2} a t^2 \), and substituting the values \( a = 588.6 \, \text{m/s}^2 \) and \( t = 0.036\, \text{s} \), the computed distance is approximately \( 0.38 \text{m} \).
The calculated distance is quite small, emphasizing how quickly airbags work, providing very little room for a safe stop under such high accelerations. It's a testament to the precision engineering behind vehicle safety measures.

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Most popular questions from this chapter

A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s\(^2\) for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50 m/s\(^2\) until it stops at the next station. Find the total distance covered.

You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top; 8.00 s later you hear the sound of the rock hitting the ground at the foot of the cliff. (a) If you ignore air resistance, how high is the cliff if the speed of sound is 330 m/s? (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain.

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